Consider a decreasing G.P. : `g_1,g_2,g_3...g_n...` Such that `g1 + g_2 + g_3 = 13` and `g_1^2+g_2^2+g_3^2=91` then which of the followng holds

Options

A. The greatest term of the G.P.is 9.
B. `3g_(4)= g_(3)`
C. `g_(1) = 1`
D. `g_(2) = 3`

Answer

To solve the problem, we need to analyze the given conditions for the decreasing geometric progression (G.P.) \( g_1, g_2, g_3, \ldots \) where:

1. \( g_1 + g_2 + g_3 = 13 \)
2. \( g_1^2 + g_2^2 + g_3^2 = 91 \)

### Step 1: Define the terms of the G.P.
Let the first term \( g_1 = a \) and the common ratio \( r \). Therefore, we can express the terms as:
- \( g_1 = a \)
- \( g_2 = ar \)
- \( g_3 = ar^2 \)

### Step 2: Set up the first equation
From the first condition, we have:
\[ g_1 + g_2 + g_3 = a + ar + ar^2 = 13 \]
Factoring out \( a \):
\[ a(1 + r + r^2) = 13 \quad \text{(Equation 1)} \]

### Step 3: Set up the second equation
From the second condition, we have:
\[ g_1^2 + g_2^2 + g_3^2 = a^2 + (ar)^2 + (ar^2)^2 = a^2 + a^2r^2 + a^2r^4 = 91 \]
Factoring out \( a^2 \):
\[ a^2(1 + r^2 + r^4) = 91 \quad \text{(Equation 2)} \]

### Step 4: Relate the two equations
From Equation 1, we can express \( a \):
\[ a = \frac{13}{1 + r + r^2} \]
Substituting this into Equation 2:
\[ \left(\frac{13}{1 + r + r^2}\right)^2 (1 + r^2 + r^4) = 91 \]
This simplifies to:
\[ \frac{169(1 + r^2 + r^4)}{(1 + r + r^2)^2} = 91 \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ 169(1 + r^2 + r^4) = 91(1 + r + r^2)^2 \]
Expanding the right side:
\[ 91(1 + 2r + r^2 + r^2 + 2r^3 + r^4) = 91(1 + 2r + 2r^2 + 2r^3 + r^4) \]
This leads to a quadratic equation in terms of \( r \).

### Step 6: Solve the quadratic equation
After simplifying, we will arrive at a quadratic equation in \( r \):
\[ 3r^2 - 10r + 3 = 0 \]
Using the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} \]
Calculating the discriminant:
\[ = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm \sqrt{64}}{6} = \frac{10 \pm 8}{6} \]
This gives us two potential solutions for \( r \):
\[ r = 3 \quad \text{or} \quad r = \frac{1}{3} \]

### Step 7: Choose the valid solution
Since the G.P. is decreasing, we choose \( r = \frac{1}{3} \).

### Step 8: Find \( a \)
Substituting \( r \) back into Equation 1:
\[ a(1 + \frac{1}{3} + \frac{1}{9}) = 13 \]
Calculating \( 1 + \frac{1}{3} + \frac{1}{9} = \frac{9 + 3 + 1}{9} = \frac{13}{9} \):
\[ a \cdot \frac{13}{9} = 13 \implies a = 9 \]

### Step 9: Determine the terms of the G.P.
Now we can find the terms:
- \( g_1 = 9 \)
- \( g_2 = 9 \cdot \frac{1}{3} = 3 \)
- \( g_3 = 9 \cdot \left(\frac{1}{3}\right)^2 = 1 \)

### Conclusion
The series is \( 9, 3, 1, \frac{1}{3}, \frac{1}{9}, \ldots \)